the potential at the center of curvature of the arc = v = Q ∕ (4πε∘a) or 
ATQ,
We have density of charge,
λ = 
Where L is the rod's length, in this case the semicircle's length L = πr
Q is the charge on the rod
The potential created at the center by an differential element of charge is:
where k is the coulomb's constant
r is the distance from dQ to center of the circle
v = ∫ 
, Where a = radius, k = 1 / 4πε∘
v = or Q ∕ (4πε∘a)
To learn more about potential from the given link
brainly.com/question/25923373
#SPJ4
Answer:
32,808.4 feet.
Explanation:
1 kilometer is equal to 3,280.84 feet.
3,280.84 feet × 10 feet = 32,808.4 feet
The answer is c 5 starrrrrr
Answer:
480.32 foot per second
Explanation:
there is a conversion thing ona famous search engine use that and I just multiple by 4
The concept required to perform this exercise is given by the coulomb law.
The force expressed according to this law is given by
Where,
q = charges of the objects
r= distance/radius
Our values are previously given, so
Replacing,
The force acting on the block are given by,
Therefore the box is accelerated upward.
